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I was fiddling with the quantum circuit used for Deutsch's algorithm and I was led to a zero state as a result which is bizarre. I don't know how to explain this result. enter image description here

***I'm using Nielsen and Chuang to study QC

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    $\begingroup$ You don't explain why you think the terms cancel out. Without that, it's hard for us to help... $\endgroup$ – DaftWullie Jul 2 at 15:25
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The terms of expression do not cancel out in the balanced function case.

We start with

$$\frac{1}{2} (|0\rangle|0 \oplus f(0)\rangle - |0\rangle|1 \oplus f(0)\rangle + |1\rangle|0 \oplus f(1)\rangle - |1\rangle|1 \oplus f(1)\rangle)$$

If $f(0) \neq f(1)$, consider the first two terms (the only ones which can cancel with each other, since the state of the first qubit is $|0\rangle$): the state of the second qubit in them is $|0 \oplus f(0)\rangle$ and $|1 \oplus f(0)\rangle$, which are different states no matter what the value of $f(0)$ is.

It will be helpful for you to consider the cases for $f(x) = x$ and $f(x) = 1-x$ separately to convince yourself that the cancellation doesn't happen.

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